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Php 提交表单后,获取另一个表中条目的ID_Php_Mysql_Forms - Fatal编程技术网

Php 提交表单后,获取另一个表中条目的ID

Php 提交表单后,获取另一个表中条目的ID,php,mysql,forms,Php,Mysql,Forms,我当前的表单将数据提交到两个不同的表中,我希望一个表中的自动递增值也存储在第二个表中 <form method="POST" action="addcocktail.php" > Cocktail Name: <input type="text" name="cocktailname" /> How To: <input type="text" name="howto" />

我当前的表单将数据提交到两个不同的表中,我希望一个表中的自动递增值也存储在第二个表中

<form method="POST" action="addcocktail.php" >
                    Cocktail Name: <input type="text" name="cocktailname" /> 
                    How To: <input type="text" name="howto" /> 
                    <br> 
                    <select id="selectingred1" name="selectingred1">
                      <?php
                      $sql = "SELECT ingredientID, name FROM tblIngredient ".
                      "ORDER BY name";

                      $rs = mysql_query($sql);

                      while($row = mysql_fetch_array($rs))
                      {
                        echo "<option value=\"".$row['ingredientID']."\">".$row['name']."</option>\n  ";
                      }
                      ?>
                    </select>
                    <select id="quantity1" name="quantity1">
                      <option></option>
                      <option>1</option>
                      <option>2</option>
                      <option>3</option>
                      <option>4</option>
                    </select>
                    <br>
<input type="submit" value="add" />
                </form>

鸡尾酒名称:
如何:

1. 2. 3. 4.
addcockbox.php:

 <?php include("databasecon.php"); ?>

<?php
mysql_select_db("mwheywood", $con);

//insert cocktail details
$sql="INSERT INTO tblCocktail (name, howto)
VALUES
('$_POST[cocktailname]','$_POST[howto]')";

$sql2="INSERT INTO tblRecipe (ingredientID, quantity)
VALUES
('$_POST[selectingred1]','$_POST[quantity1]'),
('$_POST[selectingred2]','$_POST[quantity2]'),
('$_POST[selectingred3]','$_POST[quantity3]'),
('$_POST[selectingred4]','$_POST[quantity4]')";



if (!mysql_query($sql,$con))
  {
  die('Error: you fail at life' . mysql_error());
  }
echo "cocktail added";

if (!mysql_query($sql2,$con))
  {
  die('Error: you fail at life' . mysql_error());
  }
echo "ingredients added";

mysql_close($con);

?>


执行insert查询后,如果函数成功,您可以获得insert id。

hmm我不太确定它如何知道在insert语句之后为每个选择框条目插入Cocktaild?仔细想想,您会发现。你有所有的功能,只剩下把它们放在一起了。